$\left| \begin{array}{ccc} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{array} \right| = $

$f(x) = \left| \begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{array} \right|$, then $f(100)$ is equal to:

If $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}$ is equal to

Let $A = \begin{bmatrix} 1 + x^2 - y^2 - z^2 & 2(xy + z) & 2(zx - y) \\ 2(xy - z) & 1 + y^2 - z^2 - x^2 & 2(yz + x) \\ 2(zx + y) & 2(yz - x) & 1 + z^2 - x^2 - y^2 \end{bmatrix}$. Then $\det(A)$ is equal to:

If $\theta \in \left(0, \frac{\pi}{2}\right)$,then $\left|\begin{array}{ccc} (\sin \theta+\operatorname{cosec} \theta)^2 & (\sin \theta-\operatorname{cosec} \theta)^2 & 2020 \\ (\cos \theta+\sec \theta)^2 & (\cos \theta-\sec \theta)^2 & 2020 \\ (\tan \theta+\cot \theta)^2 & (\tan \theta-\cot \theta)^2 & 2020 \end{array}\right| = $

The value of the determinant $\left| \begin{array}{ccc} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{array} \right|$ is